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In the solution of an exercise, the following reasoning is used:

$1)$ $\langle \psi_1 \psi_2 | H_1|\psi_1 \psi_2 \rangle + \langle \psi_1 \psi_2 | H_2|\psi_1 \psi_2 \rangle = \langle \psi_1 | H_1|\psi_1 \rangle \langle \psi_2 | \psi_2 \rangle + \langle \psi_1 | \psi_1 \rangle \langle \psi_2 | H_2|\psi_2 \rangle$

$2)$ $\langle \psi_1 \psi_2 | H_1|\psi_2 \psi_1 \rangle + \langle \psi_1 \psi_2 | H_2|\psi_2 \psi_1 \rangle = \langle \psi_1 | H_1|\psi_2 \rangle \langle \psi_2 | \psi_1 \rangle + \langle \psi_1 | \psi_2 \rangle \langle \psi_2 | H_2|\psi_1 \rangle$

$H_1$ and $H_2$ refer to the Hamiltonian operator.

May I ask you what property - and if possible the name of this property so that I can look it up on the internet - they used to go from the left hand side to the right hand side for both line $1)$ and $2)$?

I was unable to find anything on the internet (maybe I used the wrong terms ?), even on https://en.wikipedia.org/wiki/Bra%E2%80%93ket_notation#Properties

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What you've written from your text/exercise is an abuse of notation, but it is standard.

The composite state $|\psi \alpha\rangle$ is the tensor product of the states $|\psi\rangle \in \mathcal H_1$ and $|\alpha\rangle \in \mathcal H_2$, which is sometimes written $|\psi\alpha\rangle \equiv |\psi\rangle \otimes |\alpha\rangle$. A very typical example of such a state is a spatial wavefunction attached to a spin-1/2 state, in which case $\mathcal H_1 = L^2(\mathbb R)$ and $\mathcal H_2 = \mathbb C^2$.

The space that such composite states belong to is the tensor product of the Hilbert spaces $\mathcal H_1$ and $\mathcal H_2$, usually written $\mathcal H_1 \otimes \mathcal H_2$. The inner product on such a space can be inherited from the inner products on $\mathcal H_1$ and $\mathcal H_2$. Letting $\psi,\phi \in \mathcal H_1$ and $\alpha,\beta\in\mathcal H_2$, we have

$$\langle \psi\alpha|\phi\beta\rangle_{\mathcal H_1 \otimes \mathcal H_2} = \langle\psi|\phi\rangle_\mathcal{H_1} \cdot \langle\alpha|\beta\rangle_{\mathcal H_2}$$

Given two operators $A$ and $B$ which act on $\mathcal H_1$ and $\mathcal H_2$, we can define a new operator which acts on $\mathcal H_1\otimes \mathcal H_2$ like this:

$$\big(A \otimes B\big)\big(|\psi\rangle \otimes |\alpha\rangle\big) = \big(A|\psi\rangle\big)\otimes\big(B|\alpha\rangle\big)$$

Putting everything together (and dropping the subscripts on the inner products,

$$\langle \psi\alpha| \big(A\otimes B\big) |\phi \beta\rangle = \langle \psi|A|\phi\rangle \cdot \langle \alpha|B |\beta\rangle $$


With that out of the way, if you have a composite system made by stitching together the Hilbert spaces $\mathcal H_1$ and $\mathcal H_2$, then the Hamiltonian for the composite system is

$$H = H_1 \otimes \mathbb I_2 + \mathbb I_1 \otimes H_2$$

where $H_{1,2}$ are the Hamiltonian operators acting on $\mathcal H_1$ and $\mathcal H_2$, and $\mathbb I_{1,2}$ are the identity operators on $\mathcal H_1$ and $\mathcal H_2$. Plugging this in to what I wrote above provides you with the answer you're looking for.


More specifically, note that

$$\langle \psi_1\psi_2|H_1\otimes\mathbb I | \psi_2\psi_1 \rangle = \langle \psi_1|H_1|\psi_2\rangle \cdot \langle \psi_2|\mathbb I|\psi_1\rangle = \langle \psi_1|H_1|\psi_2\rangle \cdot \langle \psi_2|\psi_1\rangle$$

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  • $\begingroup$ I still don't really see how this applies to line 2). Why exactly does $\langle \psi_1 \psi_2 | H_1|\psi_2 \psi_1 \rangle = \langle \psi_1 | H_1|\psi_2 \rangle \langle \psi_2 | \psi_1 \rangle $ $\endgroup$ – holomorphicfunction Nov 22 '19 at 20:43
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    $\begingroup$ Line one and line two are exactly the same except for the reversed order of the states in the right-hand ket, so I'm not sure I follow your meaning. In my notation, $\langle \psi_1\psi_2|H_1\otimes\mathbb I | \psi_2\psi_1 \rangle = \langle \psi_1|H_1|\psi_2\rangle \cdot \langle \psi_2|\mathbb I|\psi_1\rangle$. $\endgroup$ – J. Murray Nov 22 '19 at 20:48
  • $\begingroup$ Aaaaah, okay, it is clear now. Thank You so much. Would it be possible to include your comment in your answer to serve as an "example" ("For example, for Line two...") ? $\endgroup$ – holomorphicfunction Nov 22 '19 at 20:50
  • $\begingroup$ After that I will accept your answer $\endgroup$ – holomorphicfunction Nov 22 '19 at 20:50
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    $\begingroup$ Okay, I've added that at the end. $\endgroup$ – J. Murray Nov 22 '19 at 20:53
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It's just notation. $H_1$ acts on Hilbert space ${\mathcal H}_1$, $H_2$ acts on Hilbert space ${\mathcal H}_2$. By $ |\psi_1,\psi_2\rangle$ they have an implicit tensor product
$$ |\psi_1,\psi_2\rangle\stackrel{\rm def}{=} |\psi_1\rangle\otimes |\psi_2\rangle\in {\mathcal H}_1\otimes {\mathcal H}_2. $$ Then physiscists, usually without saying so, extend $H_1$ and $H_2$ to the tensor product by setting $$ H_1\to H_1\otimes {\mathbb I},\quad H_2\to {\mathbb I}\otimes H_2. $$ Each factor acts on its own space, so $|\psi_2\rangle$ is invisible to the extended $H_1$, and so on.

The same ideas apply when adding angular momentum: In physics we often write $J_{\rm tot}=J_1+J_2$, but we mean $$ J_{tot}= J_1\otimes {\mathbb I}+ {\mathbb I}\otimes J_2. $$

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It would be helpful for you to define what $H_1$ and $H_2$ are in your exercise. My guess is that your source is being sloppy in their notation for how they are defining tensor products of operators. Probably, they are working on some tensored Hilbert space $\mathcal{H}_A \otimes \mathcal{H}_{B}$ and they mean something like $$ H_1 = h_1 \otimes I_{B} \\ H_2 = I_{A} \otimes h_{2} $$ where $I_{A,B}$ is the identity operator on $\mathcal{H}_{A,B}$, and $h_1$ and $h_2$ are some operators on $\mathcal{H}_A$ and $\mathcal{H}_{B}$ respectively. If $|\psi_1 \psi_2 \rangle$ means $|\psi_1 \psi_2 \rangle := | \psi_1 \rangle \otimes | \psi_2 \rangle$ then it immediately follows that $$ \langle \psi_1 \psi_2 | H_1 + H_2 |\psi_1 \psi_2 \rangle = \langle \psi_1 | h_1 |\psi_1 \rangle \; \langle \psi_2| \psi_2 \rangle + \langle \psi_1|\psi_1 \rangle \; \langle \psi_2| h_2|\psi_2 \rangle $$ which is your first identity.

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